Question
A unit vector perpendicular to the plane of a=2^i−6^j−3^k,b=4^i+3^j−^k is
Answer: Option C
:
C
a×b=∣∣
∣
∣∣^i^j^k2−6−343−1∣∣
∣
∣∣=^i(6+9)−^j(−2+12)+^k(6+24)=15^i−10^j+30^k
|a×b|=√225+100+900=35
Unit vector normal to the plane = 15^i−10^j+30^k35=3^i−2^j+6^k7
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:
C
a×b=∣∣
∣
∣∣^i^j^k2−6−343−1∣∣
∣
∣∣=^i(6+9)−^j(−2+12)+^k(6+24)=15^i−10^j+30^k
|a×b|=√225+100+900=35
Unit vector normal to the plane = 15^i−10^j+30^k35=3^i−2^j+6^k7
Was this answer helpful ?
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